Question

find the area under the cure y=sqrt(9-x) on the interval [0,5]

Answer 1

\[\int\limits_{0}^{5} \sqrt(9-x)\]

Answer 2

the square root should go over the whole (9-x)

Answer 3

how do we find the anit derivtive of that?

Answer 4

You can do this by making the algebraic substitution of

u = 9 - x

Then,

du = -dx and you can get the anti-derivative of -u^(1/2). Be careful to adjust your integration limits for the definite integral or else once you get the integral on u, convert it back to terms of "x".

Answer 5

can you show me with my number

Answer 6

np, but I type slowly! Hope you're patient!

Answer 7

yea
i need this explained so ill wait

Answer 8

thank you

Answer 9

I'm going to do the anti-derivative and then you can put in the limits.\[\int\limits_{}^{}\sqrt{9 - x} dx = - \int\limits_{}^{}\sqrt{u} du\]because u = 9 - x and du = -dx ,upon taking the differentials of both sides, so, the integrand becomes:\[- (\frac{ 2 }{ 3 }) u ^{\frac{ 3 }{ 2 }}\]Now, you can substitute the expression in "x" , which is 9 - x, back into this for "u"

Answer 10

To check this, you might want to take the derivative and you'll see that you end up with the original integrand.

Answer 11

im lost on the du part

Answer 12

You might be using different symbols and techniques than what I learned. Different authors use different symbology. The easy part is thinking of u as a function of x. Then just take the derivative of u with respect to x, which can be thought of as du/dx or u'. If you keep with du/dx, then differentials will allow that you "multiply" each side by "dx" which is just the differential of x. You probably don't use du and dx. You can also just ignore it.

Answer 13

but in my case what does du equal?

Answer 14

The kernel of the problem is just making the substitution of u = 9 - x

You can forget the theory of differentials and just concentrate on the mechanics.

And taking the derivative of what I wrote will convince you of the answer.

As for what du means, it is the differential change in u. It is similar to delta u where in the definition of the derivative we have delta u over delta x as delta x goes to 0.

Answer 15

In short, if you learned anti-derivatives without the differential, don't worry about it in the least. Just ignore where I wrote that out. Your author apparently doesn't use that symbol.

Answer 16

okayy thanks!

Answer 17

You're welcome!

Answer 18

So, with making the substitution, you get:
\[- \frac{ 2 }{ 3 }(9 - x)^{\frac{ 3 }{ 2 }}\]evaluated over the same interval.

Answer 19

yea i understand it from there
thanks!

Answer 20

np!